[proofplan]
We argue by contradiction. Suppose $\alpha \in H^k(N; \mathbb{F})$ is a non-zero class with $f^*(\alpha) = 0$. By the non-singularity of the cup product pairing on $N$, there exists a complementary class $\beta \in H^{d-k}(N; \mathbb{F})$ with $(\alpha \smile \beta)[N] \neq 0$. Pulling back through $f$ and using the naturality of the cup product and the fundamental class, we derive $\deg(f) = 0$ in $\mathbb{F}$, contradicting the hypothesis.
[/proofplan]
[step:Suppose $f^*(\alpha) = 0$ for some non-zero $\alpha$ and find a dual class $\beta$]
Assume for contradiction that there exists $\alpha \in H^k(N; \mathbb{F})$ with $\alpha \neq 0$ and $f^*(\alpha) = 0$.
Since $N$ is a compact connected $d$-dimensional $\mathbb{Z}$-oriented manifold, $H_*(N; \mathbb{F})$ is a finite-dimensional $\mathbb{F}$-vector space, hence free over $\mathbb{F}$. By the [Non-Singularity of the Cup Product Pairing](/theorems/2293) applied to $N$ with coefficients in $\mathbb{F}$, the adjoint map
\begin{align*}
H^k(N; \mathbb{F}) \to \operatorname{Hom}_\mathbb{F}(H^{d-k}(N; \mathbb{F}), \mathbb{F})
\end{align*}
is an isomorphism. Since $\alpha \neq 0$, the corresponding functional is non-zero, so there exists $\beta \in H^{d-k}(N; \mathbb{F})$ with
\begin{align*}
(\alpha \smile \beta)[N] \neq 0.
\end{align*}
By rescaling $\beta$ (we are over a field), we may assume $(\alpha \smile \beta)[N] = 1$.
[/step]
[step:Derive a contradiction by computing the degree]
Since $f: M \to N$ is a map between compact connected $\mathbb{Z}$-oriented $d$-manifolds, the induced map on top homology satisfies $f_*[M] = \deg(f) \cdot [N]$. Evaluating the degree class:
\begin{align*}
\deg(f) &= \deg(f) \cdot 1 = \deg(f) \cdot (\alpha \smile \beta)[N] = (\alpha \smile \beta)\bigl(\deg(f) \cdot [N]\bigr) \\
&= (\alpha \smile \beta)(f_*[M]) = (f^*(\alpha \smile \beta))[M] = (f^*(\alpha) \smile f^*(\beta))[M] \\
&= (0 \smile f^*(\beta))[M] = 0.
\end{align*}
The third equality uses the definition of degree. The fourth uses the naturality of the evaluation pairing: $\varphi(f_*\sigma) = (f^*\varphi)(\sigma)$ for any cochain $\varphi$ and chain $\sigma$. The fifth uses the ring homomorphism property $f^*(\alpha \smile \beta) = f^*(\alpha) \smile f^*(\beta)$. The sixth substitutes $f^*(\alpha) = 0$.
This gives $\deg(f) = 0$ in $\mathbb{F}$, contradicting the hypothesis that $\deg(f) \neq 0$ in $\mathbb{F}$.
[guided]
Let us trace through the computation more carefully. We have two manifolds $M$ and $N$, both compact, connected, $d$-dimensional, and $\mathbb{Z}$-oriented, and a continuous map $f: M \to N$ with $\deg(f) \neq 0$ in $\mathbb{F}$.
The degree is defined by $f_*[M] = \deg(f) \cdot [N] \in H_d(N; \mathbb{Z})$. Over $\mathbb{F}$, this becomes $f_*[M]_\mathbb{F} = \deg(f) \cdot [N]_\mathbb{F}$ where $[M]_\mathbb{F}$ and $[N]_\mathbb{F}$ are the images of the fundamental classes under the coefficient map $H_d(-; \mathbb{Z}) \to H_d(-; \mathbb{F})$.
Now we use the key identity: for any $\varphi \in H^d(N; \mathbb{F})$,
\begin{align*}
\varphi(f_*[M]) = (f^*\varphi)[M].
\end{align*}
This is the naturality of the Kronecker pairing with respect to continuous maps. Applying this with $\varphi = \alpha \smile \beta$:
\begin{align*}
(\alpha \smile \beta)(f_*[M]) = (f^*(\alpha \smile \beta))[M].
\end{align*}
The left side equals $(\alpha \smile \beta)(\deg(f) \cdot [N]) = \deg(f) \cdot (\alpha \smile \beta)[N] = \deg(f) \cdot 1 = \deg(f)$.
The right side equals $(f^*\alpha \smile f^*\beta)[M] = (0 \smile f^*\beta)[M] = 0$, since the cup product with the zero class is zero.
So $\deg(f) = 0$ in $\mathbb{F}$, contradicting our assumption. The contradiction arose from assuming $f^*(\alpha) = 0$ for some non-zero $\alpha$, so $f^*$ must be injective.
[/guided]
[/step]
